1. A 0.4243-g sample of CaCO 3 is dissolved in 12 M HCI and the resulting soluti

Question

1. A 0.4243-g sample of CaCO 3 is dissolved in 12 M HCI and the resulting solution is diluted to 250.0 mL in a volumetric flask.
a) how many moles of the CaCO 3 are used (formula mass = 100.1)?
b) what is the molarity of the CA 2+ in the 250 mL of solution?
c) How many moles of Ca 2+ are in a 25.00 mL aliquot of solution in lb?

2. 25.00 mL aliquots of the solution from problem 1 are titrate with EDTA to the Eriochrome Black T end point. A blank containing a small measured amount of Mg 2+ requires 3.21 mL of the EDTA to reach the end point. An aliquot to which the same amount of the Mg 2+ is added requires 24.95 mL of the EDTA to reach the end point.
a) How many milliliters of EDTA are needed to titrate the CA 2+ ion in the aliquot?
b) How many moles of the EDTA are there in the volume obtained in part A?
c) What is the molarity of the EDTA solution?

3. A 100-mL sample of hard water is titrated with the EDTA solution in Problem 2. The same amount of Mg 2+ is added as previously, and the volume of EDTA required is 31.84 mL.
a) What volume of EDTA is used in titrating the Ca 2+ in the hard water?
b) How many moles of EDTA are there in that volume?
c) How many moles of Ca 2+ are there in the 100 mL of water?
d) If the Ca 2+ comes from CaCO 3+ How many moles of CaCO 3 are there in one liter of the water? How many grams of CaCO 3 are present per liter of the water?
e) If 1 ppm CaCO 3 = 1 mg per liter, what is the water hardness in ppm CaCO 3?

Explanation / Answer

1

CaCO3 ---> Ca2+ + CO3^2-

from equation

1 mole CaCO3 = 1 mole Ca2+

a) No of moles of the CaCO3 are used = 0.4243/100.1 = 0.0042 mole

b) molarity of the Ca2+ =n/v in L= 0.0042 / 0.25 = 0.0168 M

c) moles of Ca 2+ are in a 25.00 mL = 0.0042 mole

2. Since EDTA and Ca2+ react in a 1:1 mole ratio,

   moles EDTA = moles Ca2+

a) volume of EDTA needed = 24.95-3.21 = 21.74 ml

b) No of moles of Ca2+ = Vin L*M = 0.025*0.0168 = 0.00042 mole

   moles EDTA = 0.00042 mole

c) molarity of the EDTA solution = n/v in L

      = 0.00042/0.02174 = 0.0193 M

3.

a) volume of EDTA is used in titrating the Ca 2+ in the hard water

    = 31.83-2.21 = 29.62 ml

b) moles of EDTA = M*Vin L = 0.0193*0.02962 = 0.00057 mole

c) moles of EDTA = moles of Ca 2+ are there in the 100 mL of water

   = 0.00057 mole

d) 1 mole Ca2+ = 1 mole CaCo3

= 0.00057 mole CaCo3 present in 100ml water.

e) mass of CaCO3 = 0.00057*100.1 = 0.057 grams = 57.057 mg

concentration of caco3 = mass in mg /V in L

         = 57.057/0.1 = 570.57 mg/L

water hardness in ppm CaCO3 = 570.57 ppm

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